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kozerog [31]
3 years ago
14

How were the designs of Pasteur’s and Redi’s experiments different?

Biology
1 answer:
Oksi-84 [34.3K]3 years ago
3 0
I just learned this in science a month ago! Redi used a piece of meat in a jar and covered it with a cheesecloth. He got another jar of meat and left it open. He wanted to see if magots developed from flies or not. Luis Pasteur used an 'S' tube thingy and put broth inside and boiled it. Then he waited to see if broth fogged up because of dust or microorganisms. One tube was pointed upward so dust could get inside, and the other, downward. Hope this kindof explains this topic to you!
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How big can chia plants grow? And if they get too big, do I need to put them in soil?
Paha777 [63]

Answer:

chia plants can grow 4 feet tall

water when the soil is dry

you have to grow it in soil

8 0
3 years ago
Part C Firefly luciferase is the enzyme that allows fireflies to illuminate their abdomens. Because this light generation is an
Bumek [7]

Answer:

K = 9.620 × 10⁻⁶

Explanation:

From the given information:

Temperature T= 6° C

= (273 + 6)K

= 279 K

The correct and well presentation of the reactions are:

1. Luciferin + O_2     ⇆     oxyluciferin + light   ΔG₁°

2. ATP                     ⇄   AMP + PP_i                  ΔG₂°  = -31.6 kJ/mol

The overall  ΔG° = -4.80 kJ/mol

Let's first determine the ΔG₁° for the equation (1)

ΔG° = ΔG₁° + ΔG₂°

- ΔG₁° = - ΔG° + ΔG₂°

ΔG₁° = ΔG° - ΔG₂°

ΔG₁° = ( -4.80 - (-31.6) ) kJ/mol

ΔG₁° = 26.8 kJ/mol

Using the formula:

ΔG° = -RTIn K

In \  K =\dfrac{-\Delta G^0}{RT} \\ \\ log \ K = -\dfrac{\Delta G^0}{2.303RT}

log \ K = -\dfrac{\Delta G_1^0}{2.303RT} \\ \\ log \ K = -\dfrac{26.8 \times 10^3 \ J/mol}{2.303\times 8.314 \ J/mol/K \times 279 \ K} \\ \\ log \  K =- 5.017

K = antilog (-5.017)

K = 9.620 × 10⁻⁶

6 0
3 years ago
A student made the following table comparing the support structures in animals: Examples of Support Structures in Animals Exoske
kotegsom [21]

Answer:

The exoskeleton helps support movement by working with the muscles.

Explanation:

3 0
3 years ago
Diffusion is an important process in animals and plants.
Paraphin [41]

Answer:

Diffusion helps animals in respiration. The cells in an animal's body need glucose and oxygen for respiration. Both these are carried in the blood. When blood reaches the cells the molecules of glucose and oxygen diffuse out of the blood and into the cells. As cells use up the glucose and oxygen they produce waste chemicals and carbon dioxide.

Diffusion is a very important process for photosynthesis where carbon dioxide from the stomata diffuses into the leaves and finally into the cells. This helps the plants to photosynthesize and produce more food.

Example for animals: Animals using glucose and oxygen in diffusion.

Examples for plants: Plants using carbon dioxide to photosynthesize.

8 0
3 years ago
Locomotion in Metazoa is usually due to the contraction of what? a. Skeleton b. Muscle c. Skin d. Nerves
Ede4ka [16]

Answer: b. Muscles

A metazoa is a division of animal kingdom that includes all animals expect protozoa and sponges. Metazoa division includes multicelluar animals, which exhibit highly differentiated cells. They have muscular and nerve system and well coordinated tissues and organs. Locomotion requires a coordination activity of muscular, skeletal and neural system. It is mainly achieved by contraction and relaxation of muscles that occurs due to specialized muscle proteins named as actin and myosin which receives signals from nerves.

Hence, locomotion in metazoa is due to contraction of muscles.

7 0
3 years ago
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